Ordering the two-state reference mixtures

Put $\delta :=\mu -\lambda$. The coefficient difference is $R_{\mu }-R_{\lambda }=\delta (e_{x_1}-e_{x_0})$ and has total zero. If $\delta >0$, its negative and positive parts are $\delta e_{x_0}$ and $\delta e_{x_1}$; if $\delta <0$, they are $(-\delta )e_{x_1}$ and $(-\delta )e_{x_0}$; if $\delta =0$, both are zero.

If $\delta =0$, reflexivity gives $R_{\lambda }{⪯}_s{R}_{\mu }$. If $\delta >0$, [F1] reduces this comparison to $\delta e_{x_0}⪯\delta e_{x_1}$, which follows by positive scaling of $x_0⪯x_1$ by $\delta$. Hence $\lambda <\mu$ implies $R_{\lambda }{⪯}_s{R}_{\mu }$.

If $\delta <0$ and $R_{\lambda }{⪯}_s{R}_{\mu }$, [F1] gives $c{e}_{x_1}⪯c{e}_{x_0}$ for $c:=-\delta >0$. In the ordered real field, $c^{-1}>0$: $c^{-1}\mathrm{\ne }0$, and if $-c^{-1}>0$ then closure of the positive cone under multiplication gives $c(-c^{-1})=-1>0$. But $1>0$: since $1\mathrm{\ne }0$, trichotomy gives either $1>0$ or $-1>0$; the latter would, by positive-product closure, also give $(-1{)}^2=1>0$, contradicting trichotomy. Thus $-1>0$ is impossible, and so $c^{-1}>0$. Positive scaling by $c^{-1}$ therefore implies $x_1⪯x_0$, contrary to [A1]. Thus this comparison fails when $\lambda >\mu$. The three cases prove the equivalence.